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165

Documenta Math.

The Zassenhaus Decomposition
for the Orthogonal Group:
Properties and Applications
Alexander Hahn1

Received: May 29, 2001
Communicated by Ulf Rehmann

Abstract. Zassenhaus [17] constructed a decomposition for any element in the orthogonal group of a non-degenerate quadratic space
over a field of characteristic not 2 and used it to provide an alternative description of the spinor norm. This decomposition played a
central role in the study of question of the length of an element in
the commutator subgroup of the orthogonal group with respect to
the generating set of all elementary commutators of hyperplane reflections. See Hahn [6]. The current article develops the fundamental
properties of the Zassenhaus decomposition, e.g., those of uniqueness
and conjugacy, and applies them to sharpen and expand the analysis
of [6].
2000 Mathematics Subject Classification: 20G15, 20G25, 20F05.

1. Introduction. We begin with a discussion of the length question just
mentioned. For the moment, consider any group G along with a set of generators A (not containing the identity element of G) that satisfies A−1 = A.
Of all the factorizations of an element σ ∈ G as a product of elements from A
choose one that involves the smallest number of factors. This smallest number
is defined to be the length ℓ(σ) of σ. One very basic question - necessarily in
the context of specific examples - is this: are there parameters attached to σ
from which ℓ(σ) can be read off?
A number of theorems have responded to this question. For G a Weyl group
- or more generally a Coxeter group - and A an appropriate set of hyperplane
reflections, see Humphreys [7]. Refer to Dyer [3] for a recent result in this
1 The author wishes to thank the algebraists of Louisiana State University for their splendid
organization of Quadratic Forms 2001 and their warm hospitality throughout the conference.

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context. For G a classical group and A a set of canonical elements coming from
the underlying geometry, see Hahn-O’Meara [5] for a comprehensive treatment
of the theorems of Dieudonn´e, Wall, and others. For G a classical group and
A a set of generators coming from a single conjugacy class of elements, see
Ellers-Malzan [2] and Kn¨
uppel [10]. In a related direction, interesting codes
have been constructed starting with G = SL2 (Z) and carefully selected A.
See Margulis [12, 13] and Rosenthal-Vontobel [16] for details. The connection
with the length problem is provided by the associated Cayley graph and its
diameter.
Example 1. Let G be the symmetric group on {1, . . . , n} and let A be the
set of transpositions. Let k(σ) be the the number of cycles of σ including the
trivial cycles. Then ℓ(σ) = n − k(σ).

The fact that ℓ(σ) ≤ n − k(σ) follows from the decomposition of σ into its
disjoint cycles. The other inequality is a consequence of the fact that k(στ ) =
k(σ) ± 1 for any transposition τ . A similar (but more complicated) argument
provides
Example 2. Let G be the alternating group on {1, . . . , n} and let A be the
set of three cycles, or equivalently, the set of elementary commutators of transpositions. This time, let k(σ) be the number of cycles of odd cardinality again
including the trivial cycles. Then n − k(σ) is even and ℓ(σ) = 21 (n − k(σ)).
We now turn to the orthogonal group and begin by recalling some of the basics.
For the details, see [5], especially Sections 5.2A, 5.2B, Chapter 6 (all specialized
to the orthogonal case Λ = 0) and Section 8.2A.
Let V be a non-zero, non-degenerate, n−dimensional quadratic space with
symmetric bilinear form B over a field F with char(F ) 6= 2. Denote B(x, x)
by Q(x) and 12 Q(x) by q(x). Check that B(x, y) = q(x + y) − q(x) − q(y).
Two vectors x and y are orthogonal if B(x, y) = 0. A non-zero vector x in
V that is orthogonal to itself is isotropic and it is anisotropic otherwise. A
non-degenerate plane that contains isotropic vectors is a hyperbolic plane and
an orthogonal sum of hyperbolic planes is a hyperbolic space. If U and W are
orthogonal subspaces that intersect trivially, then U ⊕ W is denoted U ⊥ W .
The orthogonal complement of a subspace U of V is denoted by U ⊥ , and the
radical of U is defined by Rad U = U ∩ U ⊥ . If W is a complement of Rad U

in U , then W is non-degenerate and U = Rad U ⊥ W is a radical splitting of
U . Any two such complements of Rad U are isometric.
Let On (V ) be the orthogonal group of V . For σ ∈ On (V ), let S be the subspace
S = (σ − 1V )V of V . This S is the space of σ. Intuitively, this is where the
”action” of σ is. In particular, there is no action on the orthogonal complement
S ⊥ of S; the fact is that S ⊥ = {x ∈ V | σ(x) = x}. Clearly, σ = 1V if and
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only if S = 0. It turns out that dim S is even if and only if σ ∈ On+ (V ), the
subgroup of On (V ) consisting of the elements of determinant 1. If η ∈ On (V )
commutes with σ, then ηS = S. We will ”transfer” properties of S to σ. For
example, σ is non-degenerate, degenerate, or totally degenerate, if S is nondegenerate, degenerate, or totally degenerate, that is, if the radical Rad S of S
is, respectively, zero, non-zero, or S. In the same way, σ is anisotropic if S is
anisotropic.
An element σ ∈ On (V ) is an involution if σ 2 = 1. It is easy to see that σ is
an involution if and only if σ = −1S . In particular, involutions have the form
S
σ = −1S ⊥ 1S ⊥ and are non-degenerate. Let v be an anisotropic vector and
define τv in On (V ) by
τv (x) = x − B(x, v)q(v)−1 v for all x ∈ V .
= −1F v . So τv is an involution.
Check that the space of τv is F v and that τv
Fv
These involutions are the hyperplane reflections or symmetries.
Theorem 1. (Cartan-Scherk-Dieudonn´e) Let G be the group On (V ) and let
A be the set of hyperplane reflections. If σ is not totally degenerate, then
ℓ(σ) = dim S. If σ is totally degenerate, then ℓ(σ) = dim S + 2.
Theorem 1 in combination with Examples 1 and 2 calls attention to the length

problem in the situation where G is the commutator subgroup Ωn (V ) of On (V )
and A the set of elementary commutators of symmetries. It seems surprising
that this question did not receive scrutiny until recently. John Hsia first called
attention to it in the case of a non-dyadic local field and it was solved in this
context in Hahn [6]. The answer is not simply a modification of the conclusion
of Theorem 1, as a comparison of Examples 1 and 2 might suggest. We will see
that, unlike Theorem 1, it depends critically on the arithmetic of the field F .
2. The Zassenhaus Decomposition. An element σ in On (V ) is unipotent
if its minimal polynomial has the form (X − 1)m for some positive integer
m. A non-trivial unipotent element is degenerate and can, therefore, exist
only if V is isotropic. The elements with minimal polynomial (X − 1)2 are
precisely the non-trivial totally degenerate elements. A degenerate element σ
with dim S = 2 is an Eichler transformation. Let S be a degenerate plane and
put S = F u ⊥ F v with u ∈ Rad S and v ∈ S. Define Σu,v ∈ On (V ) by
Σu,v (x) = x + B(x, v)u − B(x, u)v − q(v)B(x, u)u for all x ∈ V .
Then Σu,v is an Eichler transformation and all Eichler transformations have
this form. A totally degenerate Eichler transformation has minimal polynomial
(X−1)2 and one that is not totally degenerate has minimal polynomial (X−1)3 .
In particular, all Eichler transformations are unipotent.
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Let σ be any element in On (V ). Consider the subspace
X = {x ∈ V | (σ − 1V )j x = 0 some j}
of V . This unique largest space on which σ acts as a unipotent transformation
turns out to be non-degenerate. Let R = X ⊥ . Then R is non-degenerate and
X = R⊥ . Notice that σR⊥ = R⊥ . So σR = R and hence σ = σ ⊥ ⊥ σ . Put
R
R

µ = σ ⊥ ⊥ 1R and ρ = 1R⊥ ⊥ σ . Then
R

R

σ=µ · ρ
with µ unipotent and ρ non-degenerate with space R. This is the Zassenhaus
decomposition or splitting of σ. Note that µ and ρ commute.
To develop the essential properties of the Zassenhaus splitting, we need the
Wall form. Let σ ∈ On (V ). Define
( , )σ : S × S −→ F
by the equation (σx − x, σy − y)σ = B(σx − x, y) for all σx − x and σy − y
in S. This is the Wall form on S. It is non-degenerate and bilinear, but it
is almost never symmetric. In fact, ( , )σ is symmetric if and only if σ is an
involution, and in this case, (s, s′ )σ = − 21 B(s, s′ ) for all s, s′ in S. Also, ( , )σ
is alternating if and only if σ is totally degenerate.
The space S is now equipped with both the Wall form ( , )σ and the restriction
of B. When the focus is on ( , )σ , then S is denoted by Sσ . Similarly, the space
S1 of σ1 in On (V ) is written Sσ1 when ( , )σ1 is under consideration, and
analogously for σ2 . The spaces of orthogonal transformations µ, ρ, µ′ , ρ′ and

so on, will be denoted by U, R, U ′ , R′ and so on, with appropriate subscripts
when the focus is on the Wall form.
The key facts are these. Let S1 be a non-degenerate subspace of Sσ . Then
there is a unique σ1 ∈ On (V ) - the transformation belonging to S1 - such
that Sσ1 = S1 . Let S2 be the right complement of S1 in Sσ . Then S2 is
non-degenerate. If σ2 is the transformation belonging to S2 , then σ = σ1 σ2 .
Conversely, if σ = σ1 σ2 with S1 ∩ S2 = 0, then Sσ = Sσ1 ⊥ Sσ2 . This means
that the Wall forms of both Sσ1 and Sσ2 are obtained by restricting the Wall
form ( , )σ and that (s1 , s2 )σ = 0 for all s1 ∈ S1 and s2 ∈ S2 (but it is not
required that (s2 , s1 )σ = 0). For example, if σ = µρ is the Zassenhaus splitting
of σ, then because µ and ρ commute,
Sσ = U µ ⊥ R ρ = R ρ ⊥ U µ .
Another important fact asserts that elements σ and σ1 in On (V ) are conjugate
in On (V ) if and only if the spaces Sσ and Sσ1 are isometric.
To conclude this discussion of the Wall form, we note that the map
∗

∗

Θ : On+ (V ) −→ F /F 2

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∗

defined by Θ(σ) = (disc Sσ )F 2 , where disc Sσ is the discriminant of the space
Sσ , provides one of the (equivalent) definitions of the spinor norm. Its kernel
is denoted by On′ (V ). All unipotent elements are in On′ (V ). It is clear that
On′ (V ) ⊇ Ωn (V ) and it is a standard fact that if V is isotropic, then On′ (V ) =

Ωn (V ). A formula useful for computations is
Θ(σ) = Θ(ρ) = det (ρ − 1V )

R

disc R ,

where ρ is the non-degenerate component of the Zassenhaus decomposition of
∗

∗

σ and disc R ∈ F /F 2 is the discriminant of the space R relative to the form
B.
Proposition 1. Let σ = µρ be the Zassenhaus splitting of an element σ ∈
On (V ). Then S = U ⊥ R, and
i) σ is in Ωn (V ) if and only if both µ and ρ are in Ωn (V ).
ii) An element in On (V ) commutes with σ if and only if it commutes with
both µ and ρ.
Proof: Recall that On (V ) has non-trivial unipotent elements only if V is
isotropic. So any non-trivial unipotent element of On (V ) is in On′ (V ) = Ωn (V ).
This implies (i). As to (ii), observe that if η ∈ On (V ) commutes with σ, then
η stabilizes X = R⊥ . So η = η ⊥ ⊥ η and it follows that η commutes with
R
R
both µ and ρ. QED.
We next consider the question of the uniqueness of the Zassenhaus splitting. It
is not difficult to construct situations of the following sort: a non-degenerate
element σ and a non-trivial unipotent element µ0 with U0 ⊆ S such that µ0
commutes with σ and the space of ρ0 = µ−1
0 σ is S. In such a situation,
σ = 1V σ = µ0 ρ0 are two different ways of writing σ as a commuting product of
a unipotent element and a non-degenerate element. We will see that such situations are in essence the only obstruction to the uniqueness of the Zassenhaus
splitting.
Let σ = µρ be the Zassenhaus splitting of σ ∈ On (V ). Suppose that σ = µ′ ρ′
is any factorization of σ with µ′ unipotent, ρ′ non-degenerate, and such that
µ′ and ρ′ commute.
⊥

Denote by W the orthogonal complement W = R′ of the space R′ of ρ′ . By
an application of Proposition 1 (ii), the elements µ, µ′ , ρ, and ρ′ all commute
with each other. In particular, σ commutes with ρ′ . So σR′ = R′ and hence
⊥ σ ′ . The fact that ρ′
= 1W , tells us
σW = W . Therefore, σ = σ
W
R
W
′
that σ
= µ . So σ is unipotent on W and hence W ⊆ R⊥ . Therefore,
W
W
R′ = W ⊥ ⊇ R. Let T be the orthogonal complement of R in R′ . Because R′
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and R are both non-degenerate, R′ = T ⊥ R with T non-degenerate. Because
R⊥ is the largest space on which σ is unipotent, T is the largest space on which
σ ′ is unipotent. So
R

σ

= (µ

R′

T

⊥ 1R )(1T ⊥ ρ )
R

is the Zassenhaus splitting of σ ′ . Notice that 1T ⊥ ρ = ρ ′ and hence that
R
R
R
µ ⊥ 1R = µ ′ . Since µ′ and ρ′ commute with with both ρ′ and ρ, it follows
T
R
that µ′ and ρ′ stabilize the spaces R′ , R, and therefore T . So
µ

T

=σ

T

= (µ′ )(ρ′ ) .
T

T

Therefore, ρ′ is a product of two commuting unipotent transformations. So
T
ρ′ is unipotent. If T where to be non-zero, then ρ′ would fix a non-zero vector
T
in T . But this is impossible, because ρ′ is non-degenerate with space R′ . So
T = 0. Hence R′ = R and W = R⊥ . This means that µ′ ⊥ = σ ⊥ = µ ⊥
R
R
R
and hence that µ′ = µ ⊥ ⊥ µ′ . Because µ′ · ρ′ = σ = ρ , it follows
R
R
R
R
R
R
that ρ′ = 1R⊥ ⊥ (µ′ )−1 (ρ ). Therefore the obstruction to the uniqueness of
R
R
the Zassenhaus splitting is as described earlier.
Notice that U ′ ∩ R′ = U ′ ∩ R = 0 if and only if µ′ = 1R . In this case,
R
µ′ = µ and ρ′ = ρ. If R is anisotropic, then Or (R) has no non-trivial unipotent
elements, and this condition is met. The following uniqueness criterion is a
special case of our discussion.
Proposition 2. (Uniqueness) Let σ = µρ be the Zassenhaus splitting of
σ ∈ On (V ). Suppose that σ = µ′ ρ′ where µ′ is unipotent, ρ′ non-degenerate,
and S = U ′ ⊥ R′ . Then
µ′ = µ and ρ′ = ρ .
Proposition 3. (Conjugacy) Let σ and σ1 be elements in On (V ) and let
σ = µρ and σ1 = µ1 ρ1 be their Zassenhaus splittings. Then σ1 is conjugate to
σ if and only if µ1 is conjugate to µ and ρ1 is conjugate to ρ.
Proof: If σ1 is conjugate to σ then by an application of Proposition 2, µ1 is
conjugate to µ and ρ1 is conjugate to ρ. As to the converse, observe first that
Sσ = Uµ ⊥ Rρ = Rρ ⊥ Uµ and similarly for Sσ1 . If µ1 is conjugate to µ and
ρ1 is conjugate to ρ, then Uµ1 is isometric to Uµ and Rρ1 is isometric to Rρ .
Therefore Sσ1 is isometric to Sσ , and hence σ1 is conjugate to σ. QED.
3. Application to the Length Problem. Our study of the length problem
for the group Ωn (V ) and its set of generators
A = {τv τw τv τw |τv and τw non-commuting hyperplane reflections in On (V )}
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will expand on the results of Hahn [6].
Let σ ∈ Ωn (V ) be arbitrary. Note that the typical element τv τw τv τw in A is
equal to τv ττw (v) = τv τv′ where F v 6= F v ′ and Q(v) = Q(v ′ ). Conversely, any
such product is an element in A. It is a direct consequence of this fact and
Theorem 1, that
ℓ(σ) ≥

1
dim S.
2

We will therefore define σ ∈ Ωn (V ) to be short if ℓ(σ) =
ℓ(σ) > 12 dim S.

1
2

dim S and long if

Our goal is the same as that of Theorem 1, namely the complete description of
the long elements of Ωn (V ) and the determination of their lengths.
Let σ in Ωn (V ) be an involution. By an application of the Wall form, σ is short
if and only if S = W1 ⊥ · · · ⊥ Wk with dim Wi = 2 and disc Wi = 1. Totally
degenerate elements are in Ωn (V ). It follows from Theorem 1 that they are
long. We now focus on the elements in Ωn (V ) that are neither involutions nor
totally degenerate.
Theorem 2. Let σ ∈ Ωn (V ) be long with σ neither totally degenerate nor an
involution. Let σ = µρ be the Zassenhaus splitting of σ. Then
i) The space of µ satisfies U = Rad U ⊥ T with T anisotropic. The element
µ is a product of 21 (dim U ) commuting Eichler transformations, exactly
dim T of which are not totally degenerate. In particular, (µ − 1V )3 = 0.
ii) The element ρ is long and its space R is anisotropic.
iii) (Splicing Condition) The space T ⊥ R is anisotropic.
Finally, if V is isotropic, then ℓ(σ) =

1
2

dim S + 1.

Proof: In view of Hahn [6] and in particular Proposition 15, only the existence
of the factorization in (i) requires proof. By the same proposition, we know
that (µ − 1V )U ⊆ Rad U. If T = 0, then µ is totally degenerate. By HahnO’Meara [5], µ is a product of 12 (dim U ) totally degenerate commuting Eichler
transformations. So we may assume that T 6= 0. Let w1 ∈ T be non-zero.
If µw1 = w1 , then w1 is in the fixed space U ⊥ of µ. But this implies that
w1 ∈ U ∩ U ⊥ = Rad U , a contradiction. So µw1 − w1 is a non-zero vector in
Rad U . Put µw1 = u1 + w1 with u1 ∈ Rad U . Note that µ(F u1 ⊥ F w1 ) =
F u1 ⊥ F w1 and (because µ is unipotent) that the restriction of µ to this
plane has determinant 1. Let α1 = B(w1 , w1 )−1 and consider the Eichler
transformation Σu1 ,α1 w1 . Check that Σu1 ,α1 w1 (u1 ) = u1 and Σu1 ,α1 w1 (w1 ) =
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w1 + u1 . Observe that µΣ−1
u1 ,α1 w1

= 1(F u1 ⊥F w1 ) . By 8.2.16
(F u1 ⊥F w1 )
µ1 = Σ−1
u1 ,α1 w1 µ. By general facts,

of [5],

U1 ⊆
µ commutes with Σ−1
u1 ,α1 w1 . Put
(F u1 ⊥ F w1 ) + U ⊆ U . Because µ1 fixes w1 while µ does not, the fixed
space U1⊥ of µ1 strictly contains the fixed space U ⊥ of µ. It follows that
dim U1 = dim U − 2. Because µ = Σu1 ,α1 w1 · µ1 , U ⊆ (F u1 ⊥ F w1 ) + U1 .
By dimensions, U = (F u1 ⊥ F w1 ) ⊕ U1 . Since Σu1 ,α1 w1 commutes with µ,
it commmutes with µ1 . Therefore, U = (F u1 ⊥ F w1 ) ⊥ U1 . Note that
Rad U = F u1 ⊥ Rad U1 . Put U1 = Rad U1 ⊥ T1 . Because
U = (F u1 ⊥ Rad U1 ) ⊥ (F w1 ⊥ T1 )
is a radical splitting of U we know that F w1 ⊥ T1 is isometric to T and hence
that T1 is anisotropic. The element µ1 is unipotent because it is a product of
two commuting unipotent elements. An induction completes the proof. QED.
Remark: By dimension considerations, the spaces of the Eichler transformations in Theorem 2 (i) are planes with trivial intersection. Because these
Eichler transformations commute, these planes are orthogonal. Observe also
that 21 dim U ≥ dim T , and hence that dim Rad U ≥ dim T .
The next two results will show that the limitations that Theorem 2 imposes
on the components µ and ρ of the Zassenhaus splitting of a long element σ are
considerable.
Let p(X) = ak X k + · · · + a1 X + a0 be a polynomial in F [X]. We call p(X)
symmetric if the two sequences of coefficients ak , . . . , a0 and a0 , . . . , ak are
identical.
Theorem 3. Let σ ∈ Ωn (V ) be long. Then the prime decomposition of the
minimal polynomial of σ has the form
(X − 1)m p1 (X) · · · pj (X)
where 0 ≤ m ≤ 3 and the pi (X) are distinct, monic, symmetric, and irreducible.
Proof: If σ is an involution or totally degenerate, this is clear. So assume that
Theorem 2 applies to σ. Consider ρ . Because R is anisotropic, any non-zero
R
subspace W of R is non-degenerate. It follows that R = W1 ⊥ · · · ⊥ Wj where
has no non-trivial invariant subspaces.
each Wi is invariant under ρ, but ρ
Wi
By applying the results of Huppert, e.g., Satz 2.4 of [8] and Satz 4.1 of [9],
(also see the references to Cikunov in Milnor [14]), we see that the minimal
is symmetric and irreducible. QED.
polynomial of ρ
Wi

Proposition 4. Let i be the Witt index of V . Let σ ∈ Ωn (V ) be a unipotent
element with minimal polynomial (X − 1)m .
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i) If V is hyperbolic, then m ≤ 2i − 1.
ii) If V is not hyperbolic, then m ≤ 2i + 1.
In either case, there exist unipotent elements such that equality holds.
Proof: The inequalities follow by induction on i. The case i = 0 is the
anisotropic case, where we already know that 1V is the only unipotent element.
So assume that i ≥ 1. Now refer to case (2) of the proof of Theorem 2.4 of
[4] and in particular to the unipotent element τ = σY ∈ Ωn−2 (X). Let k be
the degree of the minimal polynomial of τ . Notice that V = Z ⊥ X with Z
a hyperbolic plane. So the Witt index of X is i − 1. Applying the induction
hypothesis to τ , provides the inequality k ≤ 2(i − 1) − 1 = 2i − 3 if X is
hyperbolic and k ≤ 2(i − 1) + 1 = 2i − 1 if not. It follows from the way σ and
τ are related that m ≤ k + 2. This completes the proof of the inequalities.
The construction of the required elements also follows inductively. If V is a
hyperbolic plane, then 1V is the only unipotent element and it satisfies the
equality trivially. Note next that a hyperbolic space of Witt index i contains a
non-degenerate space of Witt index i − 1 that is not hyperbolic. This implies
that it suffices to carry out the construction of the required unipotent element
in case (ii). So suppose that V is not hyperbolic with Witt index i. To get the
induction off the ground, take i = 1. Let σ = Σu,v be a non-degenerate Eichler
transformation. Then m = 3 = 2i + 1 as required. It is also easy to check
that (σ − 1V )2 V = F u. Because V is spanned by isotropic vectors, there is an
isotropic vector w in V such that (σ − 1V )2 w = u. Because i = 1, it follows
that B(σ(σ − 1V )2 w, w) = B(u, w) 6= 0.
Suppose that i ≥ 2 and let V = H ⊥ W with H a hyperbolic plane. Note
that W is not hyperbolic and that it has Witt index i − 1. For the induction
hypothesis, assume that τ is a unipotent element in Ωn−2 (W ) and that the
minimal polynomial of τ is (X − 1)k with k = 2(i − 1) + 1 = 2i − 1. Assume
further that (τ − 1W )k−1 W is a line spanned by (τ − 1W )k−1 w with w isotropic
and B(τ (τ − 1)k−1 w, w) 6= 0. Put H = F u ⊕ F v with u and v isotropic and
B(u, v) = 1. To complete the proof, we will show that σ = Σu,w · (1H ⊥ τ ) is a
unipotent element in Ωn (V ) that satisfies all the properties of τ with k + 2 in
place of k. ¿From the defining equation of Σu,w we see that σu−u = 0, σv −v =
w, and that
σx − x = τ x − x + B(τ x, w)u for all x ∈ W .
This formula and an induction shows that
(σ − 1V )j x = (τ − 1W )j x + B(τ (τ − 1W )j−1 x, w)u for all x ∈ W and j ≥ 1 .
We claim that σ has minimal polynomial (X − 1)k+2 , that (σ − 1V )k+1 V is
spanned by (σ−1V )k+1 v, and that B(σ(σ−1V )k+1 v, v) 6= 0. To see this, observe
first that (σ − 1V )k+1 x = 0 for all x ∈ W . Because (σ − 1V )u = 0, it follows
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that (σ − 1V )k+1 V is spanned by (σ − 1V )k+1 v. Recall that (σ − 1V )v = w.
Therefore,
(σ − 1V )k+1 v

= (σ − 1V )k w
= (τ − 1W )k w + B(τ (τ − 1W )k−1 w, w)u
= B(τ (τ − 1W )k−1 w, w)u 6= 0 .

Because σu = u, we see that (σ − 1V )k+2 v = 0 and hence that σ has minimal polynomial (X − 1)k+2 . Finally, B(σ(σ − 1V )k+1 v, v) = B(B(τ (τ −
1W )k−1 w, w)u, v) = B(τ (τ − 1W )k−1 w, w)B(u, v) 6= 0. The proof is complete.
QED.
The elements of Theorem 2 can be constructed as follows: Start with a long
anisotropic ρ in Ωn (V ). Choose a subspace U = Rad U ⊥ T in R⊥ such that
T ⊥ R is anisotropic and dim Rad U ≥ dim T . Split U into an orthogonal sum
of degenerate planes. For each plane choose an Eichler transformation that has
the plane as its space. Let µ be the product of these Eichler transformations and
set σ = µρ. This - by its uniqueness property - is the Zassenhaus decomposition
of σ. Therefore, the description of the long elements of Ωn (V ) has been reduced
to the following two problems:
A. Classify all long anisotropic elements ρ in Ωn (V ) and compute their
lengths in the case of an anisotropic V .
B. Determine which of the elements in Theorem 2 are actually long.
Notice that the conjugates of a long element in Ωn (V ) are long elements
of Ωn (V ). If the long element is anisotropic, then the conjugates are also
anisotropic. Thus, the problem of classifying long elements calls for the classification of their conjugacy classes.
4. Local Fields. The study of the arithmetic theory of quadratic forms
flows classically via the progression
C, R, finite fields, local fields, and global fields
from the easy situations to the hard ones. The theory over local fields makes
use of that over finite fields (via the residue class field) and the theory over
global fields - in characteristic zero these are the finite extensions of Q - is
based via local/global principles on the theory over local fields and C and R.
The benefit of hindsight, namely that the length problem that is being considered depends on the arithmetic of the field, suggests that its analysis should
proceed along the same path. Theorem 4 below is a routine application of
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Theorem 1. See Hahn [6] for the details. Observe that it applies at once to
C, R, and finite fields.
∗

∗

Theorem 4. Suppose that card F /F 2 ≤ 2. Then the totally degenerate elements σ are the only long elements in Ωn (V ), and for these ℓ(σ) = 12 dim S + 1.
Let’s turn next to the case of a local field. R+ be the set of positive real
numbers. A local field is a field F that has a valuation
| | : F −→ R+ ∪ {0}
∗

which satisfies the strong triangle inequality and with respect to which | F | is
discrete and F is complete. Let
o = {α ∈ F | |α| ≤ 1}
be the valuation ring of F and p = {α ∈ F | |α| < 1} its unique maximal
ideal. As part of the definition of local field, the residue class field o/p is
assumed to be finite. We continue the assumption that char F 6= 2. Denote
by u = {ε ∈ o | |ε| = 1} the group of invertible elements of o. Because the
maximal ideal p is principal, p = oπ for some π ∈ o. Any such π is a prime
element in o. Note that |π| is the largest value such that |π| < 1.
We refer to O’Meara [15] for the notation and the basic properties of local
fields, their quadratic forms and orthogonal groups. Two important facts about
quadratic forms over local fields are these: any non-degenerate quadratic space
of dimension five or more is isotropic, and there is, up to isometry, a unique
anisotropic four dimensional quadratic space.
It will be necessary to distinguish non-dyadic local fields from dyadic local
fields. The local field F is non-dyadic if 2 is invertible in o and dyadic if not.
So F is non-dyadic if |2| = 1 and dyadic if |2| < 1. If V is the unique 4dimensional anisotropic quadratic space, then Ω4 (V ) has index two in O4′ (V )
if F is non-dyadic, and Ω4 (V ) = O4′ (V ) if F is dyadic.
Consider a long element σ ∈ Ωn (V ) that is neither totally degenerate nor
an involution and return to the properties of the Zassenhaus decomposition
σ = µρ provided by Theorem 2. The fact that ρ is long implies that dim R ≥ 4.
Therefore,
4 ≤ dim R ≤ dim (T ⊥ R) ≤ 4 .
So T = 0, and µ is totally degenerate. In reference to Theorem 3, it follows
that the bound on m is 0 ≤ m ≤ 2. Also, dim R = 4 and R is the unique
4-dimensional anisotropic space over F . What else can be said about ρ?
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Proposition 5. Let ρ in Ωn (V ) be anisotropic with dim R = 4. Then ρ ∈
R
O4′ (R), and
i) If F is non-dyadic, then ρ is long if and only if n ≥ 5 and ρ ∈ O4′ (R) −
R
Ω4 (R) .
ii) If F is dyadic, then ρ is long if and only if ρ

R

∈ O4′ (R) = Ω4 (R) is long.

This initial answer to Question A is proved in [6]. Suppose that F is non-dyadic.
Then Proposition 5 together with Theorem 2 tell us that ℓ(σ) = 21 dim S + 1 for
any long element σ. Proposition 5 also provides a complete answer to Question
B. See Theorem 3 of [6]. It asserts that the long elements in Ωn (V ) that are not
involutions and not totally degenerate are those of Theorem 2, namely they are
precisely the elements with Zassenhaus decomposition σ = µρ, where µ = 1 V
or µ is totally degenerate and ρ satisfies (i) above. If F is dyadic, then Question
B is as yet not resolved. However, it is known that ℓ(σ) = 12 dim S + 1 for all
long elements σ.
Let V be the anisotropic 4-dimensional space over F . In view of Proposition 5
we will analyze the elements σ in O4′ (V ) that satisfy
(a) σ in O4′ (V ) − Ω4 (V ) if F is non-dyadic, and
(b) σ a long element in O4′ (V ) = Ω4 (V ) if F is dyadic.
Is there a criterion that pinpoints when an element in O4′ (V ) satisfies (a) or
(b)? A theorem of Milnor [14] tells us where to look.
Theorem 5. Let V be an n-dimensional, non-degenerate quadratic space over
a local field F . Let m(X) be a monic, irreducible polynomial in F [X] and let
deg m(X) = k. Assume that m(X) is neither X − 1 nor X + 1.
i) m(X) is the minimal polynomial of an element of On (V ) if and only if k is
n

∗

even and divides n, m(X) is symmetric, and disc V = (m(1)m(−1)) k F 2 .
ii) Given such a polynomial m(X) there is precisely one conjugacy class of
elements in On (V ) with minimal polynomial m(X).
Milnor’s result no longer holds when m(X) is reducible. Any Eichler transformation that is not totally degenerate has minimal polynomial (X − 1)3 and
provides an example showing that (i) no longer holds. The nontrivial totally
degenerate elements - all of which have minimal polynomial (X − 1)2 - show
that (ii) fails. Let µ and µ1 be totally degenerate elements. Then µ is conjugate
to µ1 if and only if their respective spaces U and U1 have the same dimension.
This follows from the conjugacy criterion given by the Wall form and the fact
Uµ and Uµ1 are both alternating.
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Let σ ∈ O4′ (V ) and let m(X) be its minimal polynomial. Milnor’s theorem suggests that it should be possible to look at m(X) and decide whether σ satisfies
condition (a) or (b) or not. In reference to Theorem 3, we are interested in
precise information about the product p1 (X) · · · pk (X). In the discussion that
follows, only the arithmetic aspects of the proofs will be provided.
Proposition 6. Let V be anisotropic with dim V = 4 and let σ ∈ O4′ (V ).
Then σ satisfies (a) or (b) if and only if σ satisfies the long criterion:
∗

Q(σx − x) = −βx2 Q(x) for all x ∈ V and some βx ∈ F .
Suppose that m(X) has a factor of the form X − a. So σx = ax for some
non-zero x in V . Because x is anisotropic, a = ±1. Since σx = x violates the
long criterion, we must have σx = −x. So a = −1. Therefore, X + 1 is the
only possible monic linear factor of m(X). If (X + 1)2 is a factor of m(X),
then −σ is a non-trivial unipotent element on some subspace of V . But this is
impossible, because V is anisotropic.
1. Suppose deg m(X) = 1. This implies that m(X) = X + 1 and hence that
σ = −1V . Because disc V = 1, −1V ∈ O4′ (V ). Check that σ = −1V
∗

satisfies the long criterion precisely when −1 ∈ F 2 .
2. Suppose deg m(X) = 2. Observe that m(X) must be irreducible. By
Theorem 5, m(X) = X 2 − cX + 1 for some c ∈ F . Notice that c 6=
±2. Every line of V contains a plane that is invariant under σ. Let
∗

W be any such plane. By Theorem 5, disc W = −(c − 2)(c + 2)F 2 .
Because V is anisotropic, W is not a hyperbolic plane, and therefore,
∗

(c − 2)(c + 2) ∈
/ F 2 . Again by Theorem 5, there is precisely one conjugacy
class of such elements σ for a given c. A spinor norm computation shows
∗

∗

that Θ(σ) = (c − 2)2 F 2 = F 2 . So any σ with minimal polynomial of this
form is in O4′ (V ). It turns out that σ satisfies the long criterion if and
∗

only if c − 2 ∈ F 2 .
3. Suppose deg m(X) = 3. By Theorem 5, m(X) is reducible. It follows
that m(X) = (X + 1)(X 2 − cX + 1) with X 2 − cX + 1 irreducible.
Again, c 6= ±2. Let p1 (X) = X + 1 and p2 (X) = X 2 − cX + 1. Put
U = p2 (σ)V and W = p1 (σ)V . Observe that U and W are planes that
are invariant under σ, that V = U ⊥ W , that σ = −1U , and that
U
has minimal polynomial X 2 − cX + 1. As in the previous case,
σ
W

∗

∗

disc W = −(c − 2)(c + 2)F 2 and (c − 2)(c + 2) ∈
/ F 2 . By Theorem 63:20
of [15], there are two isometry classes of anisotropic planes of a given
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discriminant. An application of Theorem 5 implies that there are two
conjugacy classes of σ for a given c. By a spinor norm computation,
∗

∗

Θ(σ) = −(c − 2)F 2 . So σ ∈ O4′ (V ) if and only −(c − 2) ∈ F 2 and
∗

−(c + 2) ∈
/ F 2 . The analysis of the long criterion for σ ∈ O4′ (V ) will
∗

follow shortly. We will see that if it holds, then −1 ∈ F 2 and c − 2 ∈ 4u2 .
This implies in turn that c ∈ 2u. If F is non-dyadic, the converse is true.
∗

Namely, −1 ∈ F 2 and c − 2 ∈ u2 together imply the long criterion.
4. Suppose deg m(X) = 4. In this case, either
m(X) = (X 2 − cX + 1)(X 2 − dX + 1) with distinct irreducible factors,
or
m(X) = X 4 − cX 3 − dX 2 − cX + 1 is irreducible.
The first case is very similar to case (3). The second seems complicated
and is as yet not completely understood.
We now return to case (3) and to the analysis of the long criterion. Let U˙ and
˙ denote the non-zero elements of U and W and let
W
˙ )/Q(U˙ ) .
C = Q(W
The set C is closed under multiplication by squares and hence under taking
inverses.
Assume that the long criterion holds. Applying it to U and W we get that
∗

(i)

− 1 and c − 2 are both in F 2 .

Put −1 = i2 and c − 2 = s2 and let t = −2is−1 . Applying the long criterion to
the vectors x = u + w with u ∈ U and w ∈ W , tells us that
(ii)

∗
1 + γt2
∈ F2
1+γ

for all γ ∈ C .

Conversely, the long criterion is equivalent to the combination of (i) and (ii).
We assume that (i) and (ii) hold and consider the consequences for the constant
c. We show first that t ∈ u. It follows from the discussion in paragraph 63.C of
O’Meara [15] that C contains a prime element π. Therefore C contains π i for
any odd i either positive or negative. Put t = δπ k with δ ∈ u. Taking γ = π
we get,
1 + δ 2 π 2k+1
1 + γt2
=
.
1+γ
1+π
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2k+1

π
|
If k < 0, then 2k + 1 < 0, and |1+δ|1+π|
= |π|2k+1 by the Principle of
Domination. Because 2k + 1 is odd, the element above cannot be a square.
This contradicts (ii). If k > 0, a similar contradiction is obtained by taking
γ = π −1 . Therefore, k = 0 and t ∈ u as required. Let ε = −it−1 ∈ u. Because
s = 2ε, we get

c − 2 = 4ε2 .
Therefore, c − 2 ∈ 4u2 as asserted earlier. Note that c = 2(2ε2 + 1). If F is
dyadic, then by domination, c ∈ 2u. This is also true in the non-dyadic case. If
c∈
/ u, then c ∈ p. But this would imply by Hensel’s Lemma that X 2 − cX + 1
is reducible.
∗

We now explore the converse. Assume both −1 ∈ F 2 and c − 2 ∈ 4u2 . Does
σ ∈ O4′ (V ) with such a c satisfy the long criterion, or equivalently, conditions
(i) and (ii)? Condition (i) holds trivially, so the focus is on (ii). Put c − 2 = 4ε 2
with ε ∈ u. Set s = 2ε and t = −2is−1 = −iε−1 . Notice that t ∈ u. Because C
is closed under taking inverses, (ii) is equivalent to 1 +

t2 −1
1+γ

∗

∈ F 2 for all γ in

2

c+2
c+2
C. Check that t2 − 1 = − 4+s
s2 = − c−2 = − 4ε2 . So the question is this: Is it
the case that

(iii)

1 −

∗
c+2
∈ F2
+ γ)

4ε2 (1

for all γ ∈ C?
The first step toward the answer is the observation that {|1 + γ| | γ ∈ C} is
bounded below by |4|. For suppose that |1 + γ| ≤ |4π| for some γ ∈ C. Then
∗

1 + γ = 4απ for some α ∈ o. But this means that −γ = 1 − 4απ ∈ F 2 by the
Local Square Theorem. Because C is closed under multiplication by squares,
˙ ) is not empty.
−γ ∈ C. But this implies that the intersection Q(U˙ ) ∩ Q(W
∗

∗

This would mean that V contains a plane of discriminant F 2 = −F 2 , i.e., a
hyperbolic plane. This is not possible because V is anisotropic. Now assume
that |c + 2| < |4|3 . Given the bound just established, | 4ε2c+2
(1+γ) | < |4| for all
γ ∈ C. Therefore by another application of the Local Square Theorem,
1 −

∗
c+2
2
∈
F
4ε2 (1 + γ)

for all γ in C.
We conclude the discussion of the converse by assuming that F is non-dyadic.
In this case (iii) is satisfied for any c (such that c−2 ∈ 4u2 ). Because |4| = 1, we
already know that (iii) holds when c+2 ∈ p. Since c+2 ∈ o, only the case c+2 ∈
u remains. Instead of (iii), we will verify the equivalent condition (ii). Recall
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∗

from the beginning of the analysis of case (3) that disc W = −(c − 2)(c + 2) F 2
∗

∗

and −(c + 2) ∈
/ F 2 . So disc W = −(c + 2)F 2 and, by an application of Example
∗

∗

63:15 of [15], C = πuF 2 . Let γ = πδα2 ∈ C with δ ∈ u and α ∈ F be arbitrary.
∗

If |α| ≤ 1, then |γ| < 1. So 1 + γ and 1 + t2 γ are both in F 2 by the Local
Square Theorem. Therefore (ii) holds. Suppose |α| > 1. Now |γ| > 1 and the
Local Square Theorem tells us that 1 + γ −1 and 1 + t−2 γ −1 are both squares.
−2 −1
2
γ
So 1+t
is a square. Therefore 1+γt
1+γ −1
1+γ is a square as well and (ii) holds in
this case also. The proof of the converse in the non-dyadic case is complete.
The dyadic situation is much more delicate and is not completely settled.
5. Global Fields. Let F be a global field, let V be a non-degenerate
quadratic space over F , and consider the group Ωn (V ). Not much is known
about the length question in this situation, but it is clear that local-global
considerations are relevant. Let p be a prime - Archimedean or not - and
consider the completion Vp . The first indication is the theorem that tells us
that σ ∈ Ωn (V ) if and only if σp ∈ Ωn (Vp ) for all p. Another is the fact
(analogous to what was observed in the local case) that the analysis of the
anisotropic long elements in Ωn (V ) reduces to the 4-dimensional anisotropic
long elements. This is true not only in the situation where F is a function field
or a totally complex number field (in these situations there are no anisotropic
spaces of dimension 5 or more) but in general. More precisely, if σ is an
anisotropic long element, then
σ = ω 1 · · · ω k σ1 ,
where all ωi are elementary commutators of hyperplane reflections, the space
S = W1 ⊕ · · · ⊕ Wk ⊕ S1 , and σ1 is long with dim S1 = 4.

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Alexander Hahn
Department of Mathematics
University of Notre Dame
Notre Dame, IN 46556
USA
hahn.1@nd.edu

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